# Thread: Finding and evaluating integrals!!

1. ## Finding and evaluating integrals!!

Hi Guys,
I have attached the two questions i need help with....

Really need a step by step explanation of how to complete please

2. ## Re: Finding and evaluating integrals!!

1. $\displaystyle \int (2x-3)(x+4)dx$.
Expand the product and integrate term by term.
2. $\displaystyle \int_{-2}^{-1} \frac{dx}{(x-2)^3}$
Use the substitution $\displaystyle t=x-2$ (be careful with the integration bounds!)

3. ## Re: Finding and evaluating integrals!!

For first question:

\displaystyle \begin{align*}\int (2x-3)(x+4)\,\,dx =& \int (2x^2 + 5x - 12)\,\,dx\\ =& \frac{2x^3}{3} + \frac{5x^2}{2} - 12x + C\end{align*}

You can see if the result is correct here http://www.wolframalpha.com/input/?i=integration+\int+(2x-3)(x%2B4)+dx

For Second question let $\displaystyle t = x - 2 \text{ and } \frac{dx}{dt} = 1$:

$\displaystyle \text{for lower limit } x = -2 \text { So } t = -2 -2 = -4 \text{ for upper limit } x = -1 \text{ so } t = -1 -2 = -3$

\displaystyle \begin{align*}\int_{-2}^{-1} \frac{dx}{(x-2)^3} =& \int_{-4}^{-3} \frac{dt}{t^3}\\ =& \frac{t^{-2}}{-2}\Big]_{-4}^{-3}\\ =& \frac{(-3)^{-2}}{-2} - \frac{(-4)^{-2}}{-2} \\ =& \frac{-1}{18} + \frac{1}{32} \\ =& \frac{-7}{288} \end{align*}

You can check it here: http://www.wolframalpha.com/input/?i=integration+1%2F((x-2)^3)+dx+from+x+%3D+-2+to+-1